导数的基本知识和求解方法
导数的四则运算的求导法则
求全导数的方法
解法一：直接求导法则，根据题意，a为因变量，b为自变量，则有：
a’ =(b)’(sinbtanb)+b*(sinbtanb)’=sinbtanb+b*(cosbtanb+sinbsec^2b)=sinbtanb+b(cosbtanb+sinb/cos^2b)=sinbsinb/cosb+b(cosbsinb/cosb+sinb/cos^2b)=[1/cos^2b]（sin^2bcosb+bsinbcos^2b+bsinb）=sec^2b*sinb(sinbcosb+bcos^2b+b)
解法二：已知函数变形，通过函数商的求导公式求的：a=bsinbsinb/cosb=bsin^2b/cosb然后用函数商的求导法则有：
a’=[(bsin^2b)’cosb-(bsin^2b)(cosb)’]/cos^2b =[(sin^2b+b*2sinbcosb)cosb+bsin^2b*sinb]/cos^2b=sinb*[(sinb+2bcosb)cosb+bsin^2b]/cos^2b=sinb(sinbcosb+2bcos^2b+bsin^2b)/cos^2b=sec^2b*sinb(sinbcosb+2bcos^2b+bsin^2b)=sec^2b*sinb(sinbcosb+bcos^2b+bsin^2b+bcos^2b)=sec^2b*sinb(sinbcosb+bcos^2b+b)
解法三：已知函数变形，通过求全导数的方法求得： 解：a=bsinbsinb/cosb=bsin^2b/cosbacosb=bsin^2b，两边分别求导数得到：
a’cosb+a(-sinb) =sin^2b+b*2sinbcosba’cosb-asinb=sinbsinb+2bsinbcosba’cosb=asinb+sinbsinb+2bsinbcosba’cosb=sinb(a+sinb+2bcosb)a’=sinb(a+sinb+2bcosb)/cosb =sinb(bsin^2b/cosb+sinb+2bcosb)/cosb =sinb(bsin^2b+sinbcosb+2bcosbcosb)/cos^2b=sinb(bsin^2b+bcos^2b+bcos^2b+sinbcosb)/cos^2b =sec^2b*sinb(b+bcos^2b+sinbcosb)=sec^2b*sinb(sinbcosb+bcos^2b+b)